What is the solution for 2y^{''}-16y+4=0 ?

The solution for 2y^{''}-16y+4=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}+1/4

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2y^{''}-16y+4=0

2 y^{^{''}} - 16 y + 4 = 0

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + \frac{1}{4}

Solution steps

2 y^{''} (t) - 16 y + 4 = 0

Solve linear ODE:

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + \frac{1}{4}

y = c_{1} e^{2^{1 + \frac{1}{2}} t} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + \frac{1}{4}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} + 1

y^{^{''}} + y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 9

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2

\frac{d ^{2} y}{d x ^{2}} + y = - 16 cos (3 x)

What is the solution for 2y^{''}-16y+4=0 ?
The solution for 2y^{''}-16y+4=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}+1/4